PHYS 272 Lecture Notes - Lecture 21: Resistor
23 Apr 2018
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Eight bulbs are connected in parallel to a 110 v source by two long wires of total resistance 1. 4 ohms. If a current of 240ma flows through each bulb, what is the resistance of each bulb and what fraction of the total power is wasted in each bulb. Eqv resistance 1. 4 = resistance of bulb /8. Hence resistance of bulb = 457 x 8 = 3656 ohms [ans] Total power = v x i = 110 x 0. 24 = 26. 4 w. Power in wires = i x i x r = 0. 08 w. Fraction of power in wires = 0. 08/26. 4 , which is 0. 3% [ans] Determine the equivalent resistance of the circuit below and find the voltage across each resistor. The 820 ohm and 680 ohm resistors are in parallel hence their equivalent resistance is given by. The combination is series with the 960 ohm resistor hence their resistances add up.
